I have a query that ends like this:
WHERE ID LIKE '12345%'
I want to replace the '12345' with a parameter name p_id.
If I'm useing only p_id without the % it ok but if I'm useing the % after the parameter I'm getting an error.
I'm working ORACLE database.
Any ideas?
Thanks in advance,
Roy.
try this
WHERE ID LIKE :your_parameter + "%"
|||Or, if using native Oracle vs. OLE/DB:
Where ID Like :your_parameter || '%'
Hope this helps,
Don Shelman
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